Test 3 Section 3 #9 please

When a system of linear equations has no solutions, that means the lines made by the equations in the system are parallel. Parallel lines have the same slope, so we should put both of these equations into slope-intercept form and then set up an equation to solve for k.

    \begin{align*}kx-3y&=4\\-3y&=-kx+4\\y&=\dfrac{k}{3}x-\dfrac{4}{3}\\ \\4x-5y&=7\\-5y&=-4x+7\\y&=\dfrac{4}{5}x-\dfrac{7}{5}\end{align*}

We see that one line has a slope of \dfrac{k}{3} and one has a slope of \dfrac{4}{5}. Remember, for the system to have no solutions, those slopes must be equal. So let’s solve for k.

    \begin{align*}\dfrac{k}{3}&=\dfrac{4}{5}\\k&=\dfrac{12}{5}\end{align*}

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